One-to-one function's inverse I've been trying to solve this question for a while and couldn't find the correct way. We're looking for the inverse of the given function $r$ in terms of $f^{-1}$, where $r$ is defined by:
$$r(x) = 1 - 2f(3-4x).
$$
I've tried setting $r(x)$ equal to $h(f(3-4x))$ but couldn't solve it. If you have any solution please let me know.
 A: We have the functions:


*

*$x\mapsto 3-4x$ with inverse $x\mapsto \frac34-\frac14x$. Let's denote this function by $u$.

*$x\mapsto f(x)$ with inverse $x\mapsto f^{-1}(x)$

*$x\mapsto 1-2x$ with inverse $x\mapsto\frac12-\frac12x$. Let's denote this function by $v$.


Then $r=v\circ f\circ u$ so that $r^{-1}=u^{-1}\circ f^{-1}\circ v^{-1}$.
Work out the RHS.
A: So you want to solve $$ 
y = r(x) = 1 - 2f(3-4x) $$
for $x$. We start by isolating the $f$-term, we have
$$ y = 1 - 2f(3-4x) \iff \frac{1-y}2 = f(3-4x) $$
Now apply $f^{-1}$ to both sides 
$$ \frac{1-y}2 = f(3-4x) \iff 3-4x = f^{-1}\left(\frac{1-y}2\right) $$
and isolate $x$, 
$$ 3- 4x =  f^{-1}\left(\frac{1-y}2\right) \iff x =\frac 34 - \frac 14  f^{-1}\left(\frac{1-y}2\right) $$
so $$ r^{-1}(y) = \frac 34 - \frac 14  f^{-1}\left(\frac{1-y}2\right)  $$
A: $$
r^{-1}(y)=\frac{3-f^{-1}(\tfrac{y-1}{2}{}))}{4}
$$
As a hint, substitute $y=r(x)$ and find $x$ as a function of $y$
A: To find the inverse of $r$ we let $r(x)=y$ then switch $x$ and $y$. This gives
\begin{align*}
x&=-1-2f(3-4y)\\
-\frac{x+1}{2}&=f(3-4y)\\
f^{-1}\left(-\frac{x+1}{2}\right)&=3-4y\\
y&=\frac{3-f^{-1}\left(-\frac{x+1}{2}\right)}{4}.
\end{align*}
so that 
$$r^{-1}(x)=\frac{3-f^{-1}\left(-\frac{x+1}{2}\right)}{4}.$$
A: $$ y = -1-2f(3-4 x) $$
Its inverse function is 
$$ x = -1-2f(3-4 y) $$
Just as 
$$ y = x^2/2 \rightarrow x = y^2/2. $$
Swapped re-symbolization can follow.
